TIME AND WORK

INTRODUCTION-             
Suppose Champu is given a task of constructing a building of 30 floors. He can construct 2 floors in a day. In how many days he will complete his task?

Seems Simple I guess.

Yes, 15 Days. Most of you have got it right. As he is building 2 floors in a day, in 15 days he will complete his task.
Now, does it conclude anything?????????

30(Task) = 2(Work completed per day)*15(Number of days it took to complete the task)
This takes us towards a formula;

WORK DONE=WORK RATE * TIME

BUT DO WE REQUIRE THIS FORMULA??????????????
Let's try a question.

Question-
Ram can do a piece of work in 10 days and his friend Shyam can do the same piece of work in 20 days.How long will it take for both of them to complete the work?

General Method-                 
Ram's 1-day work(work rate)=1/10
Shyam's 1-day work= 1/20
Now the question is why 1/10 or 1/20?
Why we have taken Work Done to be 1 unit?
So, it's just for simplification. We can take work as 100 also.
Now moving onto the question

Then, in one day if Ram and Shyam work together, their combined work will be =1/10 +1/20
= 3/20 units
 And we know total work to be done is 1 unit,
Hence, Time required=20/3 days ANS

Now, let us talk about some terms used in questions taking above example in consideration

EFFICIENCY-
Ram can complete the work in 10 days whereas Shyam can do in 20 days.
So, whom you will prefer for the task???
Of course, Ram it will be as he will complete the task earlier than Shyam.Hence more efficient.
But the question here arises how much Efficient is Ram as compared to Shyam????????
What are the other terminologies used?????????

Let's see some examples and after that, your TIME AND WORK questions will never be a problem.

EXAMPLE 1-A can do a piece of work in 10 days, B in 12 days and C in 15 days. They all start the
work together, but A leaves after 2 days and B leaves 3 days before the work is completed.
In how many days is the work completed?

SOLUTION-
A's 1-day work = 1/10
B's 1-day work = 1/12
C's 1-day work = 1/15
(A+B+C)'s 1 day work = 1/10 + 1/12 +1/15
Now A leaves after 2 days, means 3 of them must have worked together for 2 days
Suppose, once A left, let B and C work for 'x' days and after that C worked alone for 3 days
So, final equation,
(1/10+ 1/12 +1/15)*2 + (1/12+1/15)*x + 3*1/15 =1
Solve this equation we will get the value of x
Total days=2+x+3=x+5 will be our answer

BUT, do we need to do such cumbersome calculations????????
Suppose, Total work to be done = 60 units (LCM of 10,12,15)
Now, A's 1-day work = 60/10=6 units
          B's 1-day work = 60/12=5 units
          C's 1-day work =60/15=4 units
Now,(A+B+C)*2 + (B+C)*x + C*3=60
         15*2 + 9*x + 12=60
          9x=18
            x=2 days
Total Days=2+2+3=7 days ANS


EXAMPLE 2-
Ajay, Vijay and Sanjay are employed to do a piece of work for Rs.529. Ajay and Vijay together are
supposed to do 19/23 of the work and Vijay and Sanjay together 8/23 of the work. How much
should Ajay be paid?
(a) Rs.245 (b) Rs.295
(c) Rs.300 (d) Rs.345

SOLUTION-
Payment received depends how much work individual does.

Let total work to be done=23 units
 Work was done by Ajay + Vijay=19 units
So remaining 4 units is done by Sanjay
Likewise, Ajay completes 15 units of work
So Ajay's share = 15*529/23=15*13=Rs.345 ANS


EXAMPLE 3-
Two taps are running continuously to fill a tank. The 1st tap could have filled it in 5 hours by itself
and the second one by itself could have filled it in 20 hours. But the operator failed to realise that
there was a leak in the tank from the beginning which caused a delay of one hour in the filling of
the tank. Find the time in which the leak would empty a filled tank.

SOLUTION-
The leak is delaying the filling of the tank. Hence this leak is doing negative work.
Time is taken by first tap=5 hrs
Time is taken by second tap=20 hrs
Let the capacity of the tank be 20 units
The rate of first tap=4 units/hr
A rate of second tap=1 unit/hr
Time is taken by them when opened together=20/(4+1)=4hrs
But due to leaking its taking 1 hour more=4+1=5 hours
Capacity is 20units, time is 5 hours that means leak rate is 1 unit/hr
Hence time is taken to empty 20 units at the rate of 1unit/hr=20 hours ANS


EXAMPLE 4-
A can do as much work in 2 days as B can do in 3 days and B can do as much in 4 days as C in 5 days. A piece of work takes 20 days if all work together. How long would B
take to do all the work by himself?

SOLUTION-
Suppose A work at the rate of Aunit/day
Similarly rate of B =B unit/day
a rate of C = C unit/day
Now,2A=3B-----------(1)(work=rate*time)
        4B=5C------------(2)
multiply equation (1) by 4 equation (2) by 3
8A= 12B = 15C
Now to solve this equation lets take LCM of 8,12 and 15
i.e,120
Hence, rate of A=15unit/day
            a rate of B=10 unit/day
            a rate of C=8 unit/day
Total work=20(A+B+C)=33*20 units
Number of days B will take=33*20/10=66 days ANS


2A=3B-this sort of comparison is known as efficiency,
i.e, if B does 1 unit of work then A does 1.5 unit at the same time
Hence A is 1.5times efficient than B.


EXAMPLE 5-If 12 men and 16 women can do a piece of work in 5 days and 13 men and 24 women can do it in 4 days, how long will 7 men and 10 women take to do it?

SOLUTION-

Total work=12*5mandays + 16*5womandays---------------(1)
Total work= 13*4mandays +24*4 woman days--------------(2)
Total work must be equal, equating these two equations, we get,
60MD+80WD=52MD+96WD
8MD=16WD
1MD=2WD
putting this in equation (1) we get,
Total Work=240WD
Now, we have to find the time taken by 7M And 10W
Let it be X
Total work will still be the same and work rate of man and woman too
(7*2WD+10WD)*X=240WD
X=240/24=10DAYS ANS